Consider the following reactions Na2B4O7 2X + Y CuSO4 + Y Non-luminous flame Z + SO3 2Z + 2X + Carbon Luminous flame 2Q + Na2B4O7 + CO The oxidation states of Cu in Z and Q, respectively are :
- +2 and +1
- +1 and +1
- +1 and +2
- +2 and +2
Answer
(1) +2 and +1
Solution
This is the borax bead test. On heating, Na2B4O7 2NaBO2 + B2O3, so X = NaBO2 and Y = B2O3.
In the oxidising (non-luminous) flame, CuSO4 + B2O3 Cu(BO2)2 + SO3, so Z = Cu(BO2)2 (blue cupric metaborate): x + 2(−1) = 0 Cu = +2
In the reducing (luminous) flame, 2Cu(BO2)2 + 2NaBO2 + C 2CuBO2 + Na2B4O7 + CO, so Q = CuBO2 (colourless cuprous metaborate): x + (−1) = 0 Cu = +1
Answer: (1)